poj 1026解題報告

 

Cipher

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 13633		Accepted: 3438

Description

Bob and Alice started to use a brand-new encoding scheme. Surprisingly it is not a Public Key Cryptosystem, but their encoding and decoding is based on secret keys. They chose the secret key at their last meeting in Philadelphia on February 16th, 1996. They chose as a secret key a sequence of n distinct integers, a1 ; . . .; an, greater than zero and less or equal to n. The encoding is based on the following principle. The message is written down below the key, so that characters in the message and numbers in the key are correspondingly aligned. Character in the message at the position i is written in the encoded message at the position ai, where ai is the corresponding number in the key. And then the encoded message is encoded in the same way. This process is repeated k times. After kth encoding they exchange their message. 

The length of the message is always less or equal than n. If the message is shorter than n, then spaces are added to the end of the message to get the message with the length n. 

Help Alice and Bob and write program which reads the key and then a sequence of pairs consisting of k and message to be encoded k times and produces a list of encoded messages. 

Input

The input file consists of several blocks. Each block has a number 0 < n <= 200 in the first line. The next line contains a sequence of n numbers pairwise distinct and each greater than zero and less or equal than n. Next lines contain integer number k and one message of ascii characters separated by one space. The lines are ended with eol, this eol does not belong to the message. The block ends with the separate line with the number 0. After the last block there is in separate line the number 0.

Output

Output is divided into blocks corresponding to the input blocks. Each block contains the encoded input messages in the same order as in input file. Each encoded message in the output file has the lenght n. After each block there is one empty line.

Sample Input

10
4 5 3 7 2 8 1 6 10 9
1 Hello Bob
1995 CERC
0
0

Sample Output

BolHeol  b
C RCE

打眼一看這就是一道考察置換的題，感覺沒什麼難度，然後想都沒想就直接按模擬的方式寫了，交上去發現TLE，因爲初學沒什麼經驗，而且第一次做這方面的題，想了半天也沒有什麼好的方法，於是，上網尋求大牛們的解題報告，這才知道，原來這道題還得可能給的數據很大，需要先將每一個置換位置的週期求出來，記爲t[i]，然後進行變換時，每一個位置變換（k%t[i]）次，之後就Ok啦。。下面是我自己寫的代碼，跟大牛們的代碼相比還是有很大差距的，呵呵，只能繼續努力啦！

代碼：

語言：c++

#include<iostream> #include<cstdio> #include<cstring> using namespace std; int main() { int k,n,t[210],change[210],tt; char a1[210],a2[210]; while(cin>>n) { if(n==0) break; for(int i=0;i<n;++i) cin>>change[i]; memset(t,0,sizeof(t)); for(int i=0;i<n;++i) { t[i]=1; tt=i+1; while(change[tt-1]!=i+1) { tt=change[tt-1]; ++t[i]; } } while(cin>>k) { if(k==0) break; getchar(); cin.getline(a1,n+1); int len=strlen(a1); for(int i=len;i<n;++i) a1[i]=' '; a1[n]='/0'; strcpy(a2,a1); for(int i=0;i<n;++i) { int temp=k%t[i]; tt=i+1; for(int j=1;j<=temp;++j) tt=change[tt-1]; a2[tt-1]=a1[i]; } cout<<a2<<endl; } cout<<endl; } return 0; }